Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $n = \dfrac{x - 1}{x - 2} \times \dfrac{-2x^2 - 2x + 180}{x^2 - 10x + 9} $
Answer: First factor out any common factors. $n = \dfrac{x - 1}{x - 2} \times \dfrac{-2(x^2 + x - 90)}{x^2 - 10x + 9} $ Then factor the quadratic expressions. $n = \dfrac {x - 1} {x - 2} \times \dfrac {-2(x - 9)(x + 10)} {(x - 9)(x - 1)} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac {(x - 1) \times -2(x - 9)(x + 10) } {(x - 2) \times (x - 9)(x - 1) } $ $n = \dfrac {-2(x - 9)(x + 10)(x - 1)} {(x - 9)(x - 1)(x - 2)} $ Notice that $(x - 9)$ and $(x - 1)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac {-2\cancel{(x - 9)}(x + 10)(x - 1)} {\cancel{(x - 9)}(x - 1)(x - 2)} $ We are dividing by $x - 9$ , so $x - 9 \neq 0$ Therefore, $x \neq 9$ $n = \dfrac {-2\cancel{(x - 9)}(x + 10)\cancel{(x - 1)}} {\cancel{(x - 9)}\cancel{(x - 1)}(x - 2)} $ We are dividing by $x - 1$ , so $x - 1 \neq 0$ Therefore, $x \neq 1$ $n = \dfrac {-2(x + 10)} {x - 2} $ $ n = \dfrac{-2(x + 10)}{x - 2}; x \neq 9; x \neq 1 $